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प्रश्न
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उत्तर
'An electric bulb is rated 250 W-230V'; means that if the bulb is lighted on a 230V supply it consumes 25 ow electrical power or 25 OJ of electrical energy converts into heat and light in 1 second.
The safe Ii m it of current through the bulb is:
I = `"P"/"V" = 250/230` = 1.1 A
Current through a 60W lamp rated for 250V is:
I = `"P"/"V" = 60/250` = 0.24 A
We know that resistance of an appliance remains constant.
Resistance of the 1 amp =
R = `"P"/"I"^2 = 60/(0.24)^2 Omega`
If the line voltage falls to 200 V, power =
P = `"V"^2/"R" = (200 xx 200 xx (0.24)^2)/60` = 38.4 watt
Thus power of the lamp reduces to 38. 4 watt.
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संबंधित प्रश्न
Which of the following units could be used to measure electric charge?
(a) ampere
(b) joule
(c) volt
(d) coulomb
Three 2 V cells are connected in series and used as a battery in a circuit.
How many joules of electrical energy does 1 C gain on passing through (i) one cell (ii) all three cells?
In which direction do electrons flow?
A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate:
(i) the internal resistance and (ii) e.m.f. of the cell.
A charge of 80 C flows in a conductor for 2 minutes.
(a) Calculate the current flowing through the conductor.
(b) If the current through a heater is 4 A what charge must be passing in 8 seconds?
An auto lamp is joined to a battery of e.m.f. 4 V and internal resistance 2.5Ω. A steady current of 0.5 A flows through the circuit. Calculate the
(a) Total energy provided by battery in 10 minutes,
(b) Heat dissipated in the bulb in 10 minutes.
Answer the following question.
What is the function of a galvanometer in a circuit?
State the condition when it is advantageous to connect cells in series.
